3.66 \(\int \frac {\sin ^6(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=99 \[ \frac {\sin ^5(c+d x)}{5 a d}+\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{8 a d}-\frac {\sin (c+d x) \cos (c+d x)}{16 a d}-\frac {x}{16 a} \]

[Out]

-1/16*x/a-1/16*cos(d*x+c)*sin(d*x+c)/a/d+1/8*cos(d*x+c)^3*sin(d*x+c)/a/d+1/6*cos(d*x+c)^3*sin(d*x+c)^3/a/d+1/5
*sin(d*x+c)^5/a/d

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3872, 2839, 2564, 30, 2568, 2635, 8} \[ \frac {\sin ^5(c+d x)}{5 a d}+\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{8 a d}-\frac {\sin (c+d x) \cos (c+d x)}{16 a d}-\frac {x}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + a*Sec[c + d*x]),x]

[Out]

-x/(16*a) - (Cos[c + d*x]*Sin[c + d*x])/(16*a*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin
[c + d*x]^3)/(6*a*d) + Sin[c + d*x]^5/(5*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^6(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^6(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=\frac {\int \cos (c+d x) \sin ^4(c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{a}\\ &=\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}-\frac {\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{2 a}+\frac {\operatorname {Subst}\left (\int x^4 \, dx,x,\sin (c+d x)\right )}{a d}\\ &=\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\sin ^5(c+d x)}{5 a d}-\frac {\int \cos ^2(c+d x) \, dx}{8 a}\\ &=-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\sin ^5(c+d x)}{5 a d}-\frac {\int 1 \, dx}{16 a}\\ &=-\frac {x}{16 a}-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\sin ^5(c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.74, size = 112, normalized size = 1.13 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (120 \sin (c+d x)+15 \sin (2 (c+d x))-60 \sin (3 (c+d x))+15 \sin (4 (c+d x))+12 \sin (5 (c+d x))-5 \sin (6 (c+d x))+75 c-75 \tan \left (\frac {c}{2}\right )-60 d x\right )}{480 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(75*c - 60*d*x + 120*Sin[c + d*x] + 15*Sin[2*(c + d*x)] - 60*Sin[3*(c + d*x)]
+ 15*Sin[4*(c + d*x)] + 12*Sin[5*(c + d*x)] - 5*Sin[6*(c + d*x)] - 75*Tan[c/2]))/(480*a*d*(1 + Sec[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 70, normalized size = 0.71 \[ -\frac {15 \, d x + {\left (40 \, \cos \left (d x + c\right )^{5} - 48 \, \cos \left (d x + c\right )^{4} - 70 \, \cos \left (d x + c\right )^{3} + 96 \, \cos \left (d x + c\right )^{2} + 15 \, \cos \left (d x + c\right ) - 48\right )} \sin \left (d x + c\right )}{240 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/240*(15*d*x + (40*cos(d*x + c)^5 - 48*cos(d*x + c)^4 - 70*cos(d*x + c)^3 + 96*cos(d*x + c)^2 + 15*cos(d*x +
c) - 48)*sin(d*x + c))/(a*d)

________________________________________________________________________________________

giac [A]  time = 0.45, size = 113, normalized size = 1.14 \[ -\frac {\frac {15 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1338 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 198 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/240*(15*(d*x + c)/a + 2*(15*tan(1/2*d*x + 1/2*c)^11 + 85*tan(1/2*d*x + 1/2*c)^9 - 1338*tan(1/2*d*x + 1/2*c)
^7 - 198*tan(1/2*d*x + 1/2*c)^5 - 85*tan(1/2*d*x + 1/2*c)^3 - 15*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^
2 + 1)^6*a))/d

________________________________________________________________________________________

maple [B]  time = 0.43, size = 222, normalized size = 2.24 \[ -\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {223 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {33 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+a*sec(d*x+c)),x)

[Out]

-1/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11-17/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2
*c)^9+223/20/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7+33/20/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*
d*x+1/2*c)^5+17/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3+1/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(
1/2*d*x+1/2*c)-1/8/a/d*arctan(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 278, normalized size = 2.81 \[ \frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {85 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {198 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {1338 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {85 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {15 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a + \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) + 85*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 198*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 + 1338*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 85*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 15*sin(d
*x + c)^11/(cos(d*x + c) + 1)^11)/(a + 6*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a*sin(d*x + c)^4/(cos(d*x
+ c) + 1)^4 + 20*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6*a*sin(d*
x + c)^10/(cos(d*x + c) + 1)^10 + a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 15*arctan(sin(d*x + c)/(cos(d*x +
c) + 1))/a)/d

________________________________________________________________________________________

mupad [B]  time = 3.66, size = 106, normalized size = 1.07 \[ \frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {223\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}-\frac {x}{16\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + a/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)/8 + (17*tan(c/2 + (d*x)/2)^3)/24 + (33*tan(c/2 + (d*x)/2)^5)/20 + (223*tan(c/2 + (d*x)/2)^
7)/20 - (17*tan(c/2 + (d*x)/2)^9)/24 - tan(c/2 + (d*x)/2)^11/8)/(a*d*(tan(c/2 + (d*x)/2)^2 + 1)^6) - x/(16*a)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sin ^{6}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**6/(a+a*sec(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**6/(sec(c + d*x) + 1), x)/a

________________________________________________________________________________________